Congruations - relations
Is
$3n+2 equiv 0 bmod 4$ same as
$3n-6 equiv 0 bmod 4$ ?
I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.
modular-arithmetic congruence-relations
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
add a comment |
Is
$3n+2 equiv 0 bmod 4$ same as
$3n-6 equiv 0 bmod 4$ ?
I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.
modular-arithmetic congruence-relations
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
add a comment |
Is
$3n+2 equiv 0 bmod 4$ same as
$3n-6 equiv 0 bmod 4$ ?
I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.
modular-arithmetic congruence-relations
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
Is
$3n+2 equiv 0 bmod 4$ same as
$3n-6 equiv 0 bmod 4$ ?
I think it's the same thing because $2$ and $-6$ have the same remainder when divided with $4$.
modular-arithmetic congruence-relations
modular-arithmetic congruence-relations
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
edited Nov 29 '18 at 11:38
José Carlos Santos
151k22123224
151k22123224
asked Nov 29 '18 at 11:15
Haus
307
asked Nov 29 '18 at 11:15
Haus
307
asked Nov 29 '18 at 11:15
Haus
307
307
add a comment |
add a comment |
3 Answers
3
active
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votes
Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
add a comment |
In general: $$aequiv bmod4iff4mid a-b$$
So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
$4mid 3n+2$ and $4mid 3n-6$.
It is not difficult to prove that these statements are equivalent.
answered Nov 29 '18 at 11:31
drhab
98.2k544129
add a comment |
Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$
because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$
This law is known as the $ $ Congruence Sum Rule.
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
add a comment |
Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
add a comment |
Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
Yes, your argument is correct:begin{align}3n+2equiv3n-6pmod4&iff(3n+2)-(3n-6)equiv0pmod4\&iff8equiv0pmod4,end{align}which is true, since $4mid8$.
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
answered Nov 29 '18 at 11:21
José Carlos Santos
151k22123224
151k22123224
add a comment |
add a comment |
In general: $$aequiv bmod4iff4mid a-b$$
So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
$4mid 3n+2$ and $4mid 3n-6$.
It is not difficult to prove that these statements are equivalent.
answered Nov 29 '18 at 11:31
drhab
98.2k544129
add a comment |
In general: $$aequiv bmod4iff4mid a-b$$
So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
$4mid 3n+2$ and $4mid 3n-6$.
It is not difficult to prove that these statements are equivalent.
answered Nov 29 '18 at 11:31
drhab
98.2k544129
add a comment |
In general: $$aequiv bmod4iff4mid a-b$$
So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
$4mid 3n+2$ and $4mid 3n-6$.
It is not difficult to prove that these statements are equivalent.
answered Nov 29 '18 at 11:31
drhab
98.2k544129
In general: $$aequiv bmod4iff4mid a-b$$
So the statements $3n+2equiv 0mod4$ and $3n-6equiv 0mod4$ can be translated into
$4mid 3n+2$ and $4mid 3n-6$.
It is not difficult to prove that these statements are equivalent.
answered Nov 29 '18 at 11:31
drhab
98.2k544129
answered Nov 29 '18 at 11:31
drhab
98.2k544129
answered Nov 29 '18 at 11:31
drhab
98.2k544129
answered Nov 29 '18 at 11:31
drhab
98.2k544129
98.2k544129
add a comment |
add a comment |
Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$
because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$
This law is known as the $ $ Congruence Sum Rule.
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
add a comment |
Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$
because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$
This law is known as the $ $ Congruence Sum Rule.
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
add a comment |
Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$
because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$
This law is known as the $ $ Congruence Sum Rule.
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
Yes $!bmod m!:, color{#c00}{yequiv y'} Rightarrow, x+color{#c00}yequiv x+color{#c00}{y'}$
because $, m,mid color{#c00}{y-y'}!Rightarrow mmid x!+!y!-!(x!+!y') = color{#c00}{y-y'}$
This law is known as the $ $ Congruence Sum Rule.
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
answered Nov 29 '18 at 15:40
Bill Dubuque
209k29190629
209k29190629
add a comment |
add a comment |
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